The program should accept a name of a country as input. It should then parse the Wikipedia page for that country and find all links to the wikipedia pages of other countries on that page and make a list of all such countries.

For example, if the argument was India and http://en.wikipedia.org/wiki/India has links to wiki pages of Nepal and Pakistan, the list associated with India is [Nepal, Pakistan]. The program should recursively perform the above operation for any country you have not encountered so far.

After parsing the wiki page for India, it should parse the wiki page of Nepal, Pakistan and so on.

The final output is a json object with key being the country and value being the list of countries which were linked to the wiki page for the key country.

Sample output can be:

{"India": ["Nepal", "Pakistan"], "Nepal": ["India", "China"], "Pakistan": ["India"], "China": ["Japan"], "Japan": ["India", "China"]}

This is what I have tried, but I want to get the best approach for it:

import pycountry
import wikipedia

country_dict = {}
final_dict = {}

refr_list = []

def GetCountryDict():
    country_list = list(pycountry.countries)

    for country in country_list:
        country_dict[country.name] = 1

    return country_dict

def internalParse(internal_list,final_dict,countryDict,refr_list):

    for i in refr_list:
        if not final_dict.has_key(i):

def WikiParser(country_name,country_dict):
    wiki_resp =[]
    internal_list = []
        wiki_resp = wikipedia.page(country_name)

    for i in wiki_resp.links:

        if country_dict.has_key(i):

            if i not in internal_list:
    final_dict[country_name] = internal_list

    return internal_list, final_dict,refr_list

if __name__ == '__main__':
    country_name = raw_input("Please enter country name: ")
    countryDict = GetCountryDict()

    internal_list,final_dict,refr_list = WikiParser(country_name,countryDict)

    while refr_list:


1 Answer 1


You have quite a few violations of the style guide. This isn't necessarily crucial to follow (although highly recommended!), but more important is that your code lacks consistency.

  • You have three functions, two named with CapitalizedWords and one with mixedCase; and
  • internalParse's parameters are a mix of lower_case_with_underscores and mixedCase.

Your whitespace is at least consistent, but spaces after commas makes it easier to read; compare:

def internalParse(internal_list,final_dict,countryDict,refr_list):


def internalParse(internal_list, final_dict, countryDict, refr_list):

country_dict, final_dict and refr_list shouldn't be defined at the top level of the script; global variables are a bad idea. You could create country_dict in GetCountryDict and final_dict and refr_list in WikiParser.

The current country_dict seems a bit pointless - you fill it with 1s for some reason, but then only ever use the keys. I would suggest a set instead, and using foo in bar rather than bar.has_key(foo) (in would still work with a dictionary, by the way). This really simplifies GetCountryDict:

def get_countries():
    """Get a set of all valid country names."""
    return set(country.name for country in pycountry.countries)

Note that I have renamed the function in line with the style guide and provided a docstring to explain what it does.

internalParse has four parameters but only uses three of them; you should either remove internal_list or, better, explicitly pass it through to WikiParser (where it's currently accessed by scope).

WikiParser is currently called in two places - directly from the if __name__ == '__main__': block, and indirectly via internalParse. This makes it more difficult that it needs to be to figure out what the code is doing.

Given the output you want, a better structure might be:

if __name__ == '__main__':
    # 1. Get set of valid countries first
    valid_countries = get_countries()

    # 2. Add input validation
    country_name = input_valid_country(valid_countries)

    # 3. Call parser
    print wiki_parser(country_name, valid_countries)


  1. Already covered;
  2. See Asking the user for input until they give a valid response;
  3. Refactor wiki_parser to return the dictionary of lists that you want. You can still have something like internalParse, but as a private function that's only called via wiki_parser (conventionally, it would therefore be named with a leading underscore: _internal_parse).

I don't have pycountry or wikipedia installed, so this is untested. The ideas should be useful, though!

Given the nature of the task, one option for wiki_parser would be to do it recursively. This has advantages in terms of clear and comprehensible code, although one issue can be with hitting the system recursion limit if you delve too deeply:

def wiki_parser(country_name, valid_countries, out=None):
    """Recursively parse Wikipedia for linked countries."""
    if out is None:
        out = {}
    out[country_name] = []
    for country in _parse_page(country_name, valid_countries):
        if country not in out:
            wiki_parser(country_name, valid_countries, out=None)
    return out

This uses out both to provide the output and to determine if we've already seen a given country - membership testing (foo in bar) is as efficient with a dict as a set, as they're both hash-based (O(1), vs. O(n) for a list/tuple).

If you're wondering why I've used out=None rather than out={}, see “Least Astonishment” in Python: The Mutable Default Argument.

def _parse_page(country_name, valid_countries):
    """Parse a single page and return list of linked countries."""
        wiki_resp = wikipedia.page(country_name)
    except Exception:
        return []
    return [link for link in wiki_resp.links if link in valid_countries]

Note that except Exception: is the bare minimum - bare except is a very bad idea. Ideally, you should figure out what errors wikipedia.page can raise and handle them explicitly. I've also used a list comprehension to build the list of linked countries in one step.

You currently create a dictionary of lists, but not JSON; if you actually want to, look into json.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.